@DSquare: My point was that in general you don't technically need the chain rule. You actually could just calculate the derivative using the limit definition. Granted, that would be hard in many cases and so the chain rule is a nice tool that you can use.
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ThomasFeb 23 '14 at 19:41

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@DSquare: when you start needing the chain rule to compute $x'$, you've reached the point where your algorithm to compute derivatives no longer terminates. :p
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HurkylFeb 23 '14 at 19:42

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@DSquare: I agree that knowing how the chain rule can be extended to other non-obvious cases can be helpful in teaching the chain rule, but I also think it is helpful to teach that when finding a derivative you have different tools available. And, sure enough, the hard thing can be to choose the right tool. But I don't think that teaching that we need certain tools is helpful. Teaching this way creates the impression that only this one tool can be applied. I think it is more helpful to view the various rules as tools that you can apply to solve types of problems.
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ThomasFeb 23 '14 at 19:59

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@DSquare: I agree that the chain rule is a rule, but I disagree with saying that one must use it when doing composition. As said above, you could simply find the derivative using the limit definition. So I wouldn't state it with the imperatives must or need when talking about when to use the chain rule.
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ThomasFeb 23 '14 at 20:33

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@DSquare: Again, If I can use the limit definition to find a derivative, then I don't need to use the chain rule.
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ThomasFeb 23 '14 at 20:39

Although the question literally states "...why do you need to use chain-rule...", since the poster thought the answer should be $\frac{1}{2x-1}$, I think it's clear that the poster is struggling with "Why is there an additional factor of the derivative of the inside piece?", not whether or not you need to "use" the chain rule.

With respect to some of the other excellent answers here, running this example through the limit definition, while it does show that an additional factor appears and it is indeed the derivative of the "inside piece", I find that this explanation rarely helps students understand (or maybe the right word is believe) that this is the general rule. And frankly, running through bunch of examples or more generic limit proofs never seemed to stick with my average students either. IMHO, the shortfall here comes from us as a community being sloppy in our notation, and confusing students with generalizations and short-hand.

The poster clearly believes that the derivative of $ln(x)$ is $\frac{1}{x}$. This is a partial truth.

In fact, this is the derivative of $ln(x)$ with respect to $x$! It capitalizes on the fact that the derivative of $x$ with respect to $x$ is 1, which simplifies the rule in this case.

However, the actual rule is that the general derivative of $ln(u)$, with respect to $x$, is $\frac{1}{u}du$, where $du$ is the derivative of $u$ with respect to $x$.

If you always think of the rule as $d(ln(x)) = \frac{1}{x}dx$, your chain rule element is right there embedded in your base formulas. In fact it was there all along, we just disguised it as "wrt $x$", and then got lazy and stopped being explicit about it because "we all knew what we meant".

(I'll avoid digressing into the social commentary on why ignoring respect gets us in trouble. ;-)

This is a bit too long to be a comment, I think, but I wanted to give a different take on "Why won't it just be $\frac1{2x-1}$?"

At the time of this writing, it looks like Quincunx explained why differentiation wouldn't work out if you applied that reasoning more generally, and Thomas explained why the chain rule doesn't give that answer, but that's if you trust that the chain rule works and is applicable here (as Thomas, Cameron Williams, and Quincunx all say).

The factor of $2$ from $\frac{\mathrm d}{\mathrm dx}2x-1$ that the chain rule gives you can be seen in the line $(\star)$, where the irrelevant $-1$ has been absorbed into the "$1$" term, but the $2$'s effect can still be seen.

In the end differentiation is the calculation of the following (or a similar) limit
$$\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}.$$
This is how the derivative is defined and, as shown in Mark S.’s answer, you can actually calculate your derivative this way – if you want to.

Now, calculating every derivative this way is very tedious and that’s why general differentiation rules were deduced from the above definition and why you learn them: So you can differentiate functions without resorting to the above limit.

This starts with simple functions like $f(x)=\sin(x)+\cos(x)$. If you did not have any differentiation rules, you only could could only apply the above limit. But if you know that the derivative of a sum of functions is the sum of the individual derivatives ($(f+g)' = f' + g'$) and you know the derivatives of sine and cosine, you can calculate this derivative much more easily.

In the same way, in your original problem, the chain rule is something you can apply and that eases the task of differentiation. Without it, you probably would have to resort to calculating the above limit. In particular, there plainly is no reason to assume (or rule) that the derivative of $\ln(f(x))$ would be $\frac{1}{f(x)}$.

More general, the building of mathematical rules does not work in the way that you start with everything being allowed and then rules instruct you what to do in certain situations or forbid you to do some things. In fact, you start with everything being forbidden and rules allow you to actually do things.

Physicists and engineers use the simpler Leibniz calculus to calculate the differential quotient $\frac{dy}{dx}$ instead of using those pesky Newtonian fluxion dots ($\dot{y}$) or French apostrophes ($y'$) .. :-)

Here we used the chain rule in its guise as substition rule (or coordinate transformation $u \to x$), canceling differentials in a fraction. The important bit is the factor 2 when transforming from $u$ to $x$ coordinates.

The substituion technique is even more helpful when solving integrals.